Did the Image of a ball in $R^3$ by an invertible $3\times 3$ matrix still a ball?

Question

Let $M$ be a $3\times 3$ real symmetric invertible matrix , $r>0$ be arbitrary and $x\in\mathbb{R}^3$ a fixed vector.
Let $F=\{y\in\mathbb{R}^3\;|\;|M^{-1}x-M^{-1}y|_2\leq r\}$ where the $|\cdot|_2$ denotes the usual euclidian norm.
Then, the set $F$ is an ellipsoid centered at $x$? or a ball ? Did we have : $F=B(x,r|M|_2)$? (here $|M|_2$ denotes the 2-norm of $M$).
What I do is the following:
\begin{align} y\in F \implies M^{-1}y\in B(M^{-1}x,r)\implies y\in M(B(M^{-1}x,r))\implies y=My'\mbox{where } y'\in B(M^{-1}x,r). \end{align} It follows that \begin{align} |y-x|_2=|My'-MM^{-1}x|_2\leq |M|_2|y'-M^{-1}x|_2\leq r |M|_2 . \end{align} So we have the first inclusion.
But what about the second inclusion?
Update: the set $F$ is not a ball. It is an ellipsoid. Can we determine the equation that describes $F$? I need them because I have to evaluate an integral of a function over that domain (changes of coordinates).

Answer 1

It doesn't have to be a ball. If, say,$$M=\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix},$$$r=1$ and $x=0$, then$$F=\left\{(x_1,x_2,x_3)\in\Bbb R^3\,\middle|\,\frac{x_1^{\,2}}{a^2}+\frac{x_2^{\,2}}{b^2}+\frac{x_3^{\,2}}{c^2}\leqslant1\right\},$$which is a closed ball if and only if $a=b=c(\ne0)$.